An inequality on a matrix involving a norm and the eigenvalues

Let $A = \left( {a_{ij} } \right)$ be a matrix over $\mathbb R$, of size $n \times n$. Let $\left\{ \lambda _k \right\}_{k = 1}^n$ be the $n$ eigenvalues of the matrix. Prove the following inequality:

$$\sum_{k = 1}^n \left| {\lambda _k } \right|^2 \leqslant \sum_{i = 1}^n \sum\limits_{j = 1}^n \left| {a_{ij} } \right|^2 .$$

I can't prove it, and I'm not sure if in the problem one assumes that there exist $n$ real eigenvalues, or it's also true for complex eigenvalues. D:

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I'm fairly certain this follows from Jordan form, for example, and that one has an equality iff $A$ is diagonal over $\mathbf{C}$. Will come back to it if I can... – tkr Nov 14 '11 at 22:13
Jordan form won't quite work because RHS isn't invariant under general conjugations. – p.s. Nov 14 '11 at 22:34
Yeah, that is why I hesitated writing an answer. Glad someone wrote it out. – tkr Nov 14 '11 at 22:44
Right hand side is equal to $\operatorname{Tr}(AA^*)$ and trace is equal to the sum of the eigenvalues of the matrix. You might want to continue from there. – user13838 Nov 14 '11 at 23:03

This is in fact true even for matrices over $\mathbb{C}$. Both the left and right side are invariant under conjugating the matrix $A$ by a unitary matrix $Q$. Therefore by taking the Schur decomposition we can assume without loss of generality that $A$ is upper triangular (but potentially complex even if the original $A$ was real). Since the eigenvalues of an upper triangular matrix are just the diagonal entries, the inequality follows.